Electricity and Magnetism Exam - AP Physics C: Electricity and Magnetism

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Question

Resistors are one of the most important basic components of a circuit. With very few exceptions, all circuits have at least one kind of resistor component. An ammeter is a device that measures current flowing through a circuit. Ammeters are always connected to a circuit in series.

Which of the following accurately explains why ammeters must be connected in series within a circuit, and never in parallel?

Answer

In order to give accurate readings of current, ammeters have very low resistances. If connected in parallel, the voltage pushing current through the circuit would push a very strong current through the ammeter and virtually no current through the circuit's regular path. This would not only lead to a bad reading of current, but more often than not a broken ammeter. This phenomenon is an indication of why resistors are so important: they limit the current such that a circuit does not exceed its current-carrying capacity.

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Question

A battery is measured to have a potential of 5V. When connected to a wire with no resistors or other components, the voltage measured is 4.9V.

Why was the potential of the battery measured differently when the wire was connected?

Answer

All wires have at least some internal resistance. The most likely explanation for this is that the wire is displaying slight resistance, and therefore caused the measured potential to be less than it was before.

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Question

A capacitor and an inductor mH are connected in series to complete a simple circuit. The capacitor is initially charged to 9.9 nC.

$C=820nF,\ L=530mH$

What is the value of the maximum current that passes through the circuit?

Answer

Since there are no dissipative circuit elements (such as a resistor) present in the circuit, the energy originally held by the electric field of the capacitor oscillates between electric energy and magnetic energy over time. As such, the maximum electric energy of the capacitor is equal in value to the maximum magnetic energy of the inductor.

$E_{C_{max}}=\frac{1}{2}\frac{(Q_{max})^2}{C}$

$E_{L_{max}}=\frac{1}{2}L(I_{max})^2$

$\frac{1}{2}\frac{(Q_{max})^{2}}{C}=\frac{1}{2}L(I_{max})^{2}$

$\frac{(9.9nC)^2}{820nF}=(530mH)(I_{max})^2$

$I_{max}=1.5*10^{-5}A$

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Question

Consider a spherical shell with radius and charge . What is the magnitude of the electric field at the center of this spherical shell?

Answer

According to the shell theorem, the total electric field at the center point of a charged spherical shell is always zero. At this point, any electric field lines will result in symmetry, canceling each other out and creating a net field of zero at that point.

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Question

A parallel plate capacitor has a capacitance of . If the plates are apart, what is the area of the plates?

$\epsilon_0=8.85* 10^{-12} \frac{F}{m}$

Answer

The relationship between capacitance, distance, and area is $C=\epsilon_0 \frac{A}{d}$ . We can rearrange this equation to solve for area.

$A=\frac{Cd}{\epsilon_0}$

Now, we can use the values given in teh question to solve.

$A=\frac{1F(110^{-2}m)}{8.85 10^{-12} \frac{F}{m}}=1.1*10^9m^2$

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Question

Charge is distributed uniformly over the area of the two plates of a parallel plate capacitor, resulting in a surface area charge density of $\pm \sigma$ on the plates (the top plate is positive and the bottom is negative, as shown below). Each plate has area and are separated by distance . A material of dielectric constant $\kappa_{material}$ has been placed between the two plates.

Which of the following would not result in an increase in the measure of electric potential difference between the two plates?

Answer

The electric potential difference created between the plates of a parallel plate capactor is given by the equation:

$V = \frac{Q}{C}$

The charge can be calculated by using the equation:

$Q=\sigma A$

The value of the capacitance is related to the dimensions of the capacitor with the equation:

$C=\kappa \frac{\varepsilon _{o}A}{d}$

Combining these equations yields:

$V=\frac{\sigma A}{\frac{\kappa \epsilon_0A}{d}}=\frac{\sigma d}{\varepsilon_{o} \kappa }$

The area becomes inconsequential, while the potential is directly proportional to the surface charge density and the distance between the plates, and inversely proportional to the dielectric of the material between the plates. Changing the area does not cause any change in the potential difference measured between the plates, and changing any of the other variables would cause a resultant change in the potential difference.

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Question

A straight copper wire has a fixed voltage applied across its length. Which of the following changes would increase the power dissipated by this wire?

Answer

Relevant equations:

$R = \frac{\rho L}{A}$

Current and resistance are inversely proportional to one another, assuming voltage is fixed. Since , changes in current effect the power more than changes in resistance do. Thus, we need current to increase, meaning that resistance must decrease.

To decrease resistance, we could:

1. Change the material of the wire to one of lesser resistivity

2. Decrease the length of the wire

3. Increase the cross-sectional area of the wire

4. Decrease the temperature of the wire (very slight effect on resistance)

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Question

A battery is measured to have a potential of 5V. When connected to a wire with no resistors or other components, the voltage measured is 4.9V.

If the current through the wire is measured to be 2A, how much thermal energy is being lost per second as soon as the wire is connected to the battery?

Answer

First, we must know that the wire has some internal resistance . To calculate this, we need to know the potential drop through the wire, which must be the difference we saw from the initial voltage reading to the second. This value, 0.1V, we plug into Ohm's law to calculate the resistance of the wire.

$R=0.05\Omega$

The question asks for energy lost per second; this value is equivalent to the power.

Use our values to solve.

$P=(2A)^2(0.05\Omega)=0.2W$

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Question

A simple circuit contains two $12\Omega$ resistors in parallel, connected to a 20V source. What power is being provided by the source to the circuit?

Answer

The power supplied to the circuit can be calculated using the equation:

$P=IV=\frac{V^2}{R}$

To use this equation, we need to find the equivalent resistance of the circuit. Use the equation for equivalent resistance in parallel:

$\frac{1}{R_{eq}}=\frac{1}{R_1}+\frac{1}{R_2}$

$\frac{1}{R_{eq}}=\frac{1}{12\Omega}+\frac{1}{12\Omega}=\frac{1}{6\Omega}$

$R_{eq}=6\Omega$

Now that we have the resistance and the voltage, we can solve for the power.

$P=\frac{V^2}{R}=\frac{(20V)^2}{6\Omega}\approx 67W$

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Question

Three resistors R1, R2, and R3 and the capacitor, C, are connected to an ideal battery V to complete the circuit as shown.

After the circuit has been connected for a very long time, the currents in each branch of the circuit are measured to be some values , , and , and the charge on the capacitor some value .

Which of the following equations is a correct expression regarding the voltage of the circuit after a long time?

Answer

Identifying the sum of the voltage drops and rises (Kirchoff's Loop Law) around the three possible loops of this circuit is the key to answering this question correctly. The following signs can be assigned to each of the circuit elements based on the direction of the currents given.

Use Ohm's law and the equation for capacitance to derive terms for the voltage across each element of the circuit.

$C=\frac{Q}{V}\rightarrow V=\frac{Q}{C}$

There are three possible paths through the circuit, resulting in three correct equations that could be derived:

$0=+V-\frac{Q}{}C-R_{3}I_{3} -R_{1}I_{1}$

$0=-\frac{Q}{}C-R_{3}I_{3} +R_{2}I_{2}$

$0=+V-R_{2}I_{2} -R_{1}I_{1}$

Only one of the given answer options matches up correctly to these.

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Question

Three resistors R1, R2, and R3 and the capacitor, C, are connected to an ideal battery V to complete the circuit as shown.

After the circuit has been connected for a very long time, the currents in each branch of the circuit are measured to be some values , , and , and the charge on the capacitor some value .

Which of the following equations is a correct expression regarding the voltage of the circuit after a long time?

Answer

Identifying the sum of the voltage drops and rises (Kirchoff's Loop Law) around the three possible loops of this circuit is the key to answering this question correctly. The following signs can be assigned to each of the circuit elements based on the direction of the currents given.

Use Ohm's law and the equation for capacitance to derive terms for the voltage across each element of the circuit.

$C=\frac{Q}{V}\rightarrow V=\frac{Q}{C}$

There are three possible paths through the circuit, resulting in three correct equations that could be derived:

$0=+V-\frac{Q}{}C-R_{3}I_{3} -R_{1}I_{1}$

$0=-\frac{Q}{}C-R_{3}I_{3} +R_{2}I_{2}$

$0=+V-R_{2}I_{2} -R_{1}I_{1}$

Only one of the given answer options matches up correctly to these.

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Question

A straight copper wire has a fixed voltage applied across its length. Which of the following changes would increase the power dissipated by this wire?

Answer

Relevant equations:

$R = \frac{\rho L}{A}$

Current and resistance are inversely proportional to one another, assuming voltage is fixed. Since , changes in current effect the power more than changes in resistance do. Thus, we need current to increase, meaning that resistance must decrease.

To decrease resistance, we could:

1. Change the material of the wire to one of lesser resistivity

2. Decrease the length of the wire

3. Increase the cross-sectional area of the wire

4. Decrease the temperature of the wire (very slight effect on resistance)

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Question

A battery is measured to have a potential of 5V. When connected to a wire with no resistors or other components, the voltage measured is 4.9V.

If the current through the wire is measured to be 2A, how much thermal energy is being lost per second as soon as the wire is connected to the battery?

Answer

First, we must know that the wire has some internal resistance . To calculate this, we need to know the potential drop through the wire, which must be the difference we saw from the initial voltage reading to the second. This value, 0.1V, we plug into Ohm's law to calculate the resistance of the wire.

$R=0.05\Omega$

The question asks for energy lost per second; this value is equivalent to the power.

Use our values to solve.

$P=(2A)^2(0.05\Omega)=0.2W$

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Question

A simple circuit contains two $12\Omega$ resistors in parallel, connected to a 20V source. What power is being provided by the source to the circuit?

Answer

The power supplied to the circuit can be calculated using the equation:

$P=IV=\frac{V^2}{R}$

To use this equation, we need to find the equivalent resistance of the circuit. Use the equation for equivalent resistance in parallel:

$\frac{1}{R_{eq}}=\frac{1}{R_1}+\frac{1}{R_2}$

$\frac{1}{R_{eq}}=\frac{1}{12\Omega}+\frac{1}{12\Omega}=\frac{1}{6\Omega}$

$R_{eq}=6\Omega$

Now that we have the resistance and the voltage, we can solve for the power.

$P=\frac{V^2}{R}=\frac{(20V)^2}{6\Omega}\approx 67W$

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Question

An infinitely long wire has a current of running through it. Calculate the magnetic field at a distance away from the wire.

$\mu_0=4\pi *10^{-7}\frac{Tm}{A}$

Answer

For infinitely long wires, the formula for the magnetic field is $B=\frac{\mu_0I}{2\pi r}$ , where is the current and is the distance from the wire.

The magnetic field is calculated using our given values.

$B=\frac{(4\pi10^{-7}\frac{\text{Tm}}{\text{A}})(3\text{A})}{2\pi(0.02\text{m})}$

$B=310^{-5}\ \text{T}$

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Question

A solenoid is long and it is made up of turns of wire. How much current must run through the solenoid to generate a magnetic field of inside of the solenoid?

$\mu_0=4\pi*10^{-7}\frac{Tm}{A}$

Answer

The formula for the magnetic field inside the solenoid is $B=\frac{N}{L}\mu_0I$ , where is the number of turns of wire, is the length of the solenoid, and is the current.

We want to find the current so we solve for .

$I=\frac{BL}{N\mu_0}$

Plug in the values.

$I=\frac{(1\text{T})(0.5\text{m})}{(800)(4\pi*10^{-7}\frac{\text{Tm}}{\text{A}{}})}$

$I=497\ \text{A}$

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Question

Two parallel wires a distance apart each carry a current , and repel each other with a force per unit length. If the current in each wire is doubled to , and the distance between them is halved to $\frac{R}{2}$ , by what factor does the force per unit length change?

Answer

Relevant equations:

$\oint B\cdot dl = \mu _o I$

$F = I L \times B$

Step 1: Find the original and new magnetic fields created by wire 1 at wire 2, using Ampere's law with an Amperian loop of radius or $\frac{R}{2}$ , respectively.

Original

$B (2\pi R) = \mu _o I \rightarrow B = \frac{\mu _o I}{2 \pi R}$

New

$B (2\pi (\frac{R}{2})) = \mu _o (2I) \rightarrow B = \frac{\mu _o (2I)}{2\pi \frac{R}{2}} = \frac{4\mu _o I}{2\pi R}$

Since the wires are parallel to each other and wire 1's field is directed circularly around it, in each case wire 1's field is perpendicular to wire 2.

Step 2: Find the original and new magnetic forces per unit length on wire 2, due to the field created by wire 1.

$F = IL \times B \rightarrow \frac{F}{L} = IB$

Original

$\frac{F}{L} = I * \frac{\mu _o I }{2 \pi R} = \frac{\mu _o I^2}{2\pi R}\textup{}$

New

$\frac{F}{L} = 2I * \frac{4\mu _o I}{2\pi R} = \frac{8 \mu _o I^2}{2 \pi R}$

So, the new force per unit length is 8 times greater than the original.

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Question

A region of uniform magnetic field, , is represented by the grey area of the box in the diagram. The magnetic field is oriented into the page.

A stream of protons moving at velocity $\vec{v}$ is directed into the region of the magnetic field, as shown. Identify the correct path of the stream of protons after they enter the region of magnetic field.

Answer

The magnetic force on a moving charged particle is given by the equation:

$\vec{F}=q,\vec{v}\times \vec{B}$

Isolating the directional component of this equation yields the understanding that the resulting force on a moving charged particle is perpendicular to the plane of the velocity vector and magnetic field vector. Using the right-hand-rule on this cross-product shows that the velocity vector right-crossed into the magnetic field vector into the page yields a magnetic force vector upward on a positive charge. This will result in a semi-circular path oriented vertically upward.

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Question

A metal ring is placed in a uniform magnetic field perpendicular to the plane of the ring. An emf of magnitude 15V is induced around the ring by increasing the field through it from zero to 5mT at a constant rate. If the area enclosed by the ring is $1 *10^{-2}m^2$ , what is the time interval over which the field is increased?

Answer

Relevant equations:

$\epsilon = \frac{-\Delta \Phi}{\Delta t} = \frac {- \Delta (B\cdot A)}{\Delta t}$

Given:

$A = 1 * 10^{-2} m^2$

$\Delta B = 5mT = 5 * 10^{-3}T$

$\epsilon = 15 V$

Plug in (disregarding direction of emf):

$15 V= \frac{(5 * 10^{-3}T)(1 * 10^{-2}m^2)}{\Delta t}$

$\Delta t = \frac{(510^{-3}T)(1 10^{-2}m^2)}{15V}= 3.33 * 10^{-6}s = 3.33\mu s$

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Question

A lamp has a bulb. If the house wiring provides to light up that bulb, how much current does the bulb draw?

Answer

The formula for power is ,and we are given the following values.

$P=60\ \text{W}$

$V=110\ \text{V}$

Solve for the current, .

$I=\frac{P}{V}=\frac{60\ \text{W}}{110\ \text{V}}=0.55\ \text{A}$

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